ÌâÄ¿ÄÚÈÝ

ÏÂÁÐʵÑéÖУ¬ÒÀ¾ÝʵÑé²Ù×÷¼°ÏÖÏ󣬵óöµÄ½áÂÛÕýÈ·µÄÊÇ
 
²Ù×÷
ÏÖ¡¡Ïó
½á¡¡ÂÛ
A£®
²â¶¨µÈŨ¶ÈµÄNa2CO3ºÍNa2SO3 ÈÜÒºµÄpH
Ç°ÕßpH±ÈºóÕߵĴó
·Ç½ðÊôÐÔ£º
B£®
ÎÞÉ«ÈÜÒºÖеμÓÂÈË®ºÍCCl4£¬
Õñµ´¡¢¾²ÖÃ
ϲãÈÜÒºÏÔ×ÏÉ«
Ô­ÈÜÒºÖÐÓÐ
C£®
ÏòÈÜÒºXÖмÓÈëÏ¡ÑÎËᣬ²¢½«²úÉúµÄÎÞÉ«ÆøÌåͨÈë³ÎÇåʯ»ÒË®ÖÐ
Éú³É°×É«³Áµí
ÈÜÒºXÖÐÒ»¶¨º¬ÓÐ
»ò
D£®
ÏòijÎÞÉ«ÈÜÒºÖеμÓÏõËáËữµÄBaCl2ÈÜÒº
²úÉú°×É«³Áµí
ÈÜÒºÖÐÒ»¶¨º¬ÓÐ
 
B

ÊÔÌâ·ÖÎö£ºAÏîÖÐÒª±È½Ï·Ç½ðÊôÐÔ£º£¬¿ÉÒԱȽÏÁ½ÕßËù¶ÔÓ¦µÄË᣺ÁòËáºÍ̼ËáµÄÇ¿Èõ£¬´íÎó¡£BÏîÕýÈ·¡£CÖв»ÄÜÅųýÑÇÁòËá¸ùµÄ¸ÉÈÅ£¬´íÎó¡£DÏî²»ÄÜÅųýÑÇÁòËá¸ùµÄ¸ÉÈÅ£¬´íÎó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÑÇÂÈËáÄÆ£¨NaClO2£©ÊÇÒ»ÖÖ¸ßЧÑõ»¯¼Á¡¢Æ¯°×¼Á¡£ÒÑÖª£ºNaClO2±¥ºÍÈÜÒºÔÚζȵÍÓÚ38¡æʱÎö³öµÄ¾§ÌåÊÇNaClO2¡¤3H2O£¬¸ßÓÚ38¡æʱÎö³ö¾§ÌåÊÇNaClO2£¬¸ßÓÚ60¡æʱNaClO2·Ö½â³ÉNaClO3ºÍNaCl¡£ÀûÓÃÏÂͼËùʾװÖÃÖƱ¸ÑÇÂÈËáÄÆ¡£

Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©×°ÖâÚÖвúÉúClO2µÄ»¯Ñ§·½³ÌʽΪ           ¡£×°Öâ۵Ä×÷ÓÃÊÇ             ¡£
£¨2£©´Ó×°Öâܷ´Ó¦ºóµÄÈÜÒº»ñµÃNaClO2¾§ÌåµÄ²Ù×÷²½ÖèΪ£º
¢Ù¼õѹ£¬55¡æÕô·¢½á¾§£»¢Ú          £»¢Û       £»¢ÜµÍÓÚ60¡æ¸ÉÔµÃµ½³ÉÆ·¡£
£¨3£©×¼È·³ÆÈ¡ËùµÃÑÇÂÈËáÄÆÑùÆ·10gÓÚÉÕ±­ÖУ¬¼ÓÈëÊÊÁ¿ÕôÁóË®ºÍ¹ýÁ¿µÄµâ»¯¼Ø¾§Ì壬ÔÙµÎÈëÊÊÁ¿µÄÏ¡ÁòËᣬ³ä·Ö·´Ó¦£¨ClO2£­+ 4I£­+4H+ ¡ú2H2O+2I2+Cl£­£©¡£½«ËùµÃ»ìºÏÒºÅä³É250mL´ý²âÈÜÒº¡£ÅäÖÆ´ý²âÒºÐèÓõ½µÄ¶¨Á¿²£Á§ÒÇÆ÷ÊÇ            £»
£¨4£©È¡25.00mL´ý²âÒº£¬ÓÃ2.0 mol/L Na2S2O3±ê×¼ÒºµÎ¶¨£¨I2 +2S2O32£­¡ú2I£­+S4O62£­£©£¬ÒÔµí·ÛÈÜÒº×öָʾ¼Á£¬´ïµ½µÎ¶¨ÖÕµãʱµÄÏÖÏóΪ          ¡£Öظ´µÎ¶¨2´Î£¬²âµÃNa2S2O3ÈÜҺƽ¾ùֵΪ20.00 mL¡£¸ÃÑùÆ·ÖÐNaClO2µÄÖÊÁ¿·ÖÊýΪ             ¡£
£¨5£©Í¨¹ý·ÖÎö˵Ã÷×°ÖâÙÔÚ±¾ÊµÑéÖеÄ×÷Óà                              ¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø